Definite integral (2x+3)/(x²+3x+2) from 0 to 1

This problem asks to evaluate the definite integral of a rational function on the interval [0, 1].

Problem

$\int_{0}^{1} \frac{2 x + 3}{x^{2} + 3 x + 2} d x$

Solution

Step 1

First, factor the denominator. The quadratic x² + 3x + 2 has roots x = −1 and x = −2, therefore:

Why do we factor the denominator?

Factoring the denominator into linear factors is the first step when integrating rational fractions. Knowing the factors, we can represent the fraction as a sum of partial fractions, each easily integrated. Without factoring, the method of undetermined coefficients cannot be applied.

x^{2} + 3 x + 2 = (x + 1) (x + 2)

Step 2

The integrand is a proper rational fraction (degree of the numerator 1 is less than degree of the denominator 2). Decompose it into partial fractions:

How do we decompose a proper fraction into partial fractions?

Each linear factor (x − a) in the denominator corresponds to a partial fraction of the form A/(x − a). If a factor is repeated k times, it corresponds to k fractions. In our case, the denominator has two distinct linear factors — (x + 1) and (x + 2), so the decomposition has the form A/(x + 1) + B/(x + 2).

How do we find coefficients A and B?

The method of undetermined coefficients is used. Bring the sum to a common denominator and equate the numerators. We obtain A(x+2) + B(x+1) = 2x + 3. Substitute the roots: at x = −1 we get A = 1; at x = −2 we get −B = −1, so B = 1.

\frac{2 x + 3}{(x + 1) (x + 2)} = \frac{A}{x + 1} + \frac{B}{x + 2}

Step 3

Equate the numerators after bringing to a common denominator:

A (x + 2) + B (x + 1) = 2 x + 3

Substituting the roots of the denominator, we find: A = 1, B = 1.

Step 4

Substitute the decomposition into the integral:

\int_{0}^{1} (\frac{1}{x + 1} + \frac{1}{x + 2}) d x

Step 5

Integrate each term and apply the Newton-Leibniz formula:

[_{0}^{1} ln | x + 1 | + ln | x + 2 |

Step 6

Substitute the limits. ln 1 = 0, so we get ln 2 + ln 3 − ln 2 = ln 3.

Answer

$ln 3$

Frequently asked questions

How do I evaluate a definite integral of a rational function?

You need to: (1) if the fraction is improper — extract the polynomial part, (2) factor the denominator, (3) decompose the proper fraction into partial fractions, (4) find the coefficients by the method of undetermined coefficients, (5) integrate each term, (6) substitute the limits of integration using the Newton-Leibniz formula.

What is the method of undetermined coefficients?

It is a method of finding coefficients A, B, C… in the decomposition of a proper fraction into partial fractions. The idea: bring the sum to a common denominator, equate the numerators, and then equate coefficients of equal powers of x. An alternative is to substitute specific values of x (for example, the roots of the denominator) into the identity.

Why do logarithms appear in the answer?

Because after decomposition into partial fractions, we obtain integrals of the form ∫ A/(x − a) dx = A ln|x − a| + C. This is a standard situation when integrating rational functions with linear factors in the denominator.

What if the denominator has a repeated root?

If a linear factor (x − a) is repeated k times, it corresponds to a sum of k partial fractions: A₁/(x − a) + A₂/(x − a)² + … + Aₖ/(x − a)ᵏ. For example, for (x + 1)² the decomposition is A/(x + 1) + B/(x + 1)². This is shown in the next problem.